# C* Algebra

1. Aug 21, 2005

### Hurkyl

Staff Emeritus
What's a good intro text on (noncommutative) C* Algebras?

2. Aug 21, 2005

### mathwonk

3. Aug 22, 2005

### matt grime

If you have the money then ROe and HIgson wrote a very long and very expensive book on them. i don't know what's in it but if it's anything like their lectures (which it almost certanily wil be since we were effectively taught from the draught of the book) then it will be comprehensive covering all you could ever want to know. blackadar's book is good shold you find you want to do more with the subject but isn't an introduction.

try www.math.psu.edu/roe or the same with higson instead of roe and poke around, maybe you can get something from their teaching stuff. oh and they call the study of C* algebras geometric functional analysis

4. Aug 22, 2005

### mathwonk

one of my friends, a specialist in the topic for 30 years, says:

William Arveson has a pretty basic book, called just C*-algebras, I think,
although it is a bit old fashioned. Ken Davidson has a more modern book
called C*-algebras by Example, which is quite good.

5. Aug 23, 2005

### matt grime

That's it, C* algebras by example, that's the one we used for the course.

6. Aug 23, 2005

### Hurkyl

Staff Emeritus
Yay, library at work had "An Invitation to C*-Algebras" (Arveson's book). "C*-Algebras by Example" was checked out, though.

Last edited: Aug 23, 2005
7. Aug 23, 2005

### fourier jr

gillman/jerison's rings of continuous functions is apparently a classic but i haven't looked at it very closely because the stuff is still slightly over my head.

8. Aug 23, 2005

### Hurkyl

Staff Emeritus
All right, I have some questions already.

I'm looking at the spectrum A^ of a commutative C*-algebra A which is given on page 2 as being the set of all nonzero complex homomorphisms of A, with its usual topology. (It seems to say it's treating A as a Banach algebra)

I'm presuming a complex homomorphism is a continuous homomorphism into C. What exactly is the "usual topology" on the set of such things?

It then considers C(A^) as the continus maps A^ → C vanishing at ∞. What does it mean to vanish at ∞?

Oh wait, I found my first question on Wikipedia: a net fλ of elements in A^ converges to f iff it converges pointwise. It will take a bit of work to really understand that, but at least I know what it is now!

Last edited: Aug 23, 2005
9. Aug 23, 2005

### mathwonk

well who knows what usual means. there are several standard topologies on a family of functions , pointwise convergence is the most naive and often least ineteresting as it implies very little, so perhaps not the one meant.

uniform convergence on compact sets is nice, and convergence of all sets of values is typical and weak. there is weak convergence and weak star convergence and so on. good luck.

we should probaly ask my friend the expert again but I hate to do so too often. maybe i'll ask a different expert.

10. Aug 23, 2005

### mathwonk

vanishing at infinity usually means approaching zero off compact subsets, i.e. for every epsilon there exists a compact subset off which the value is less than epsilon.

but that definition is for locally compact spavces which most banach spaces are not so maybe off bounded sets, instead of off compact sets.

Last edited: Aug 23, 2005
11. Aug 23, 2005

### Hurkyl

Staff Emeritus
Oh, BTW, Wikipedia says the "weak-* topology". I'm happier with nets than weak topologies, so I'll work with that version for now!

Arveson says that A^ is locally compact, so your definition should be applicable.

Should that be "off closed, bounded sets" in your guess for arbitrary Banach spaces?

12. Aug 23, 2005

### Hurkyl

Staff Emeritus
Oh, I get it.

If I let elements of A act on elements of A^ via:

x&omega; := &omega;(x)

It's just saying that each x in A is then a continuous map A^ &rarr; C, right?

13. Aug 23, 2005

### Hurkyl

Staff Emeritus
Bleh, that's just what the weak-* topology says! So either I really am happy with the weak-* topology, or I was tricked into being happy with nets!!!

14. Aug 24, 2005

### mathwonk

the usual topology is norm topology, i.e. the norm of a complex homomokrphism is its sup on the unit ball.

15. Aug 25, 2005

### mathwonk

Banach Algebra Techniques in Operator Theory, Vol. 179
Ronald*G.*Douglas

16. Aug 26, 2005

### Hurkyl

Staff Emeritus
I think I might have to check out a reference on banach algebras if I want any hope of working through An Invitation to C*-Algebras... I'll check to see if the library has that one.

So, I guessed that ||&omega;|| might mean its operator norm, and had proved the statement that gave me pause (that &omega; must have norm 1).

And I finally got through the rest of the proof, having only to take one item on faith: for a self-adjoint x in A, its norm as an element of C(A^), which I presume is the sup norm, and is thus the sup of |&omega;(x)| for all &omega;, is equal to $\lim_n ||x^n||^{1/n}$.

But then another question surfaces! I know what the spectrum of an operator acting on a Hilbert space is, but the text is speaking mainly of the spectrum with respect to a Banach Algebra...

Specifically, if x is an element of the Banach Algebra A, then there's something called $\textbox{sp}_A(x)$, the spectrum of x with respect to A.

Any ideas on what that is? I had a guess as to what it might be, but I think I'm wrong because I was unable to work out something the book said should take "a moment's thought".

17. Aug 26, 2005

### matt grime

sp_A(x) is the set of complex numbers y for whcih x-yI is not invertible, i'd guess. this isnt' quite the same thing as being an eigenvalue. for instance if we take the trasformation of l^2 the space of square summable sequences that sends x_n to (x_n)/n then 0 is in its spectrum since it is not invertible but is not an eigenvalue.

and the other statement is easy enouigh to prove (about norm being the sup of that thing) and if i recall correctly (ie i don't know ti off the opt of my head exactly so it isnt' obvious, but the proof is easy to understand when you read it. should appear in any functional analysis/hilbert space book worth its salt).

Last edited: Aug 26, 2005
18. Aug 27, 2005

### mathwonk

as matt said a linear map can be noninvertible by being either not bijective or not bounded below, and there are also divisions of the "specxtrum" into pure point spectrum and continuous spectrum although these distinctions may be different.

a basic tool is path integration around parts of the spectrum, as i recall from a course i did not follow 40 years ago, based on the book of edgar lorch, spectral theory.